CS 5104 course notes March 22, 2006

Transcription

1 CS 5104 course notes March 22, 2006 The Halting Problem A TM = { M, w M is a TM and M accepts w } U = On input M, w, where M is a TM and w is a string: 1. Simulate M on input w. 2. If M ever enters its accept state, accept; if M ever enters its reject state, reject. [ U recognizes A TM, but does not decide it, because if M loops forever, so does U ] [ A TM is not decidable, but how do we prove it? ] Diagonalization Definition A set A is countable if it is finite or there is a one-to-one correspondence between all the elements of A and N [ If there is a one-to-one correspondence between all the elements of any two sets, we say they have the same cardinality (or size) ] [ show that the even numbers are countable ] [ show that the rational numbers are countable ] R is uncountable It s sufficient to show that [0, 1] is uncountable. Let f: N [0, 1] be one-to-one and onto. [ one-to-one: f(47) and f(635) can t map to the same real number ] [ onto: every real is included in the mapping ] f(1) = 0.b 1,1 b 1,2 b 1,3 b 1,4 b 1,5 f(2) = 0.b 2,1 b 2,2 b 2,3 b 2,4 b 2,5 f(3) = 0.b 3,1 b 3,2 b 3,3 b 3,4 b 3,5 f(4) = 0.b 4,1 b 4,2 b 4,3 b 4,4 b 4,5 f(5) = 0.b 5,1 b 5,2 b 5,3 b 5,4 b 5,5 where each b i,j is a binary digit (0 or 1)

2 CS 5104 course notes March 22, 2006 We construct a real number a = 0.a 1 a 2 a 3 that is not included in this mapping. a 1 b 1,1 ( if b 1,1 is 0, a 1 is 1; if b 1,1 is 1, a 1 is 0 ) a 2 b 2,2 a 3 b 3,3 a 4 b 4,4 a 5 b 5,5 Suppose a is in the mapping. Then f(n) = a for some n. The n-th digit in f(n) is b n,n The n-th digit in a is a n But by construction a n b n,n [ why can t we have 1 = , 2 = , 3 = , ] A TM is undecidable ( recall that A TM = { M, w M is a TM and M accepts w } ) Suppose A TM is decidable Let H be a decider for A TM " accept Then H = reject if M accepts w if M rejects or loops on w Construct D = On input M : [ M is a TM ] 1. Run H on input M, M [ ex: Pascal compiler written in Pascal ] 2. Output the opposite of what H outputs (if H accepts, reject; if H rejects, accept) Running H on input D, D yields a contradiction: Case A: H accepts D, D ( meaning that D accepts D ) Therefore we reject ( meaning D rejects D ) Case B: H rejects D, D ( meaning that D rejects D ) Therefore we accept ( meaning D accepts D ) In both case, we get a contradiction, therefore A TM is not decidable. [ the book shows how this proof can be viewed as a diagonalization proof ]

3 CS 5104 course notes March 22, 2006 Definition A language is co-turing-recognizable if its complement in Turingrecognizable. A language is decidable iff it is Turing-recognizable and co-turing-recognizable. ( ) Assume A is decidable Then L is Turing-recognizable And L is decidable So L is Turing-recognizable Therefore, A is decidable A and A are both Turing-recognizable ( ) Assume both A and A are Turing-recognizable Let M 1 be a TM that recognizes A Let M 2 be a TM that recognizes A Construct M = On input w: 1. Run both M 1 and M 2 on input w in parallel 2. If M 1 accepts, accept; if M 2 accepts, reject w A M 1 halts & accepts M halts & accepts w A M 2 halts & rejects M halts & rejects Therefore, M decides A Therefore, A & A are Turing-recognizable A is decidable Corollary A TM is not Turing-recognizable If it were, ATM would be decidable (which is isn t)

4 CS 5104 course notes March 22, 2006 Reducibility HALT TM = { M, w TM M halts on input w } is undecidable Suppose HALT TM is decidable Let R be a decider for HALT TM HALT TM is undecidable 1. Run R on M, w 2. If R rejects (M does not halt on w), reject 3. If R accepts (M halts on w), run M on w 4. If M accepts, accept 5. If M rejects, reject E TM = { M M is a TM and L(M) = } is undecidable Suppose E TM is decidable Let R be a decider for E TM E TM is undecidable 1. Construct M 1 that rejects all strings that are not w, and accepts w only if M accepts w. ( M1 = On x: if x w, reject else Run M on w; if M accepts, accept ) [ M 1 is not a decider] [ we are not running it, we are merely constructing it ] 2. Run R on M 1 3. R rejects M 1 L(M 1 ) M 1 accepts w M accept w; accept M, w 4. R accepts M1 L(M 1 ) = M 1 does not accepts w M does not accept w (it reject or loops on w); reject M, w REGULAR TM = { M M is a TM and L(M) is regular } is undecidable Suppose REGULAR TM is decidable Let R be a decider for REGULAR TM REGULAR TM is undecidable

5 CS 5104 course notes March 22, Construct M 2 that accepts all string in the non-regular language 0 n 1 n, and accepts all other string only if M accepts w. [ therefore if M accepts w, M 2 recognizes Σ*, which is regular ] ( M 2 = On x: if x has form 0 n 1 n, accept else Run M on w; if M accepts, accept ) [ M 2 is not a decider] [ we are not running it, we are merely constructing it ] 6. Run R on M 2 7. R rejects M 2 L(M 2 ) is regular M 2 accepts all strings M accepts w; accept M, w 8. R accepts M1 L(M 1 ) is not regular M 2 only accepts string of form 0 n 1 n M does not accept w (it reject or loops on w); reject M, w EQ TM = { M 1, M 2 L(M 1 ) = L(M 2 ) } is undecidable ( show that if EQ TM is decidable, so is E TM ) [ fairly easy ] ALL CFG = { G G is a CFG and L(G) = Σ* } [ proof is in book; non-trivial ] The Domino Problem (PCP) [ describe the domino problem, state that its undecidable ] " a % A single domino: ' A set of dominos: (" b % ca& ', " a % ', " ca % a & ', " abc % + ) c & ', * - Problem: write a program that list the dominos (repeats OK) so that: top string of symbols = bottom string of symbols (if such a listing exists) " a %" For example: ' b %" ca & ' ca %" a & ' a %" ' abc % c & ' is a solution to the set above. Impossible! [ Not that it takes to long you can t do it on a computer ] [ next week : mapping reducibility ]